Question

A 25.0-mL aliquot of an aqueous quinine solution was diluted to 50.0 mL and found to have an absorbance of 0.656 at 348 nm when measured in a 2.50-cm cell.A second 25.0-mL aliquot was mixed with 10.00 mL of a solution containing 25.7 ppm of quinine; after dilution to 50.0 mL, this solution had an absorbance of 0.976 (2.50-cm cell). Calculate the concentration of quinine in parts per million in the sample.

Answer 1

The value is
`C =  21.07 ppm`

Step-by-step explanation:

From the question we are told that

The volume of the first aliquot is
`V_a1  =  25.0 \  mL`

The volume after dilution is
`V_A =  50.0mL`

The Absorbance of the first solution is
`A_1 =  0.656`

The wavelength at which the absorbance occurred is
`\lambda _1 =  348 nm  =  348*10^(-9) \  m`

The length of the cell
`l  = 2.50 \  cm`

The volume the second aliquot is
`V_(a2)  =  25.0 \  mL`

The volume of the solution it was mixed with is
`V_s  =  10.00mL`

The concentration of quinine in the 10 mL solution
`Z =  25.7ppm`

The volume of the second solution after dilution is
`V_d = 50.0 mL`

The absorbance of the second solution is
`A_2 =  0.976`

Generally the concentration of quinine is mathematically represented as

`C =  (A_1 *Z *  V_s  )/((A_2 -A_1) V_(a2))`

=>
`C =  ( 0.656  *  25.7 *  10.0)/( (0.976 -  0.656) *  25)`

=>
`C =  21.07 ppm`